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A mass attached to two horizontal springs, vertical motion 
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#1
Mar611, 09:41 PM

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1. The problem statement, all variables and given/known data
A mass is connected between two identical springs with the same spring constant, k. The equilibrium length of each spring is L, but they are stretched to 2L when the mass is in equilibrium. Ignoring gravity, find the equations of motion and then angular frequency of oscillation, when the mass is displaced by a small distance +/x.   ______m______   this is what the system looks like, with each string 2L in the picture. The x direction is vertical 2. Relevant equations F=kx a(x)=(w^2)(x) 3. The attempt at a solution from examining the forces I got that a(x)=(2kxsin(theta))/m i didnt know what to do next would w=squareroot(2k/m) 


#2
Mar711, 05:03 AM

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AM 


#3
Mar711, 06:20 PM

P: 70

here is an attached picture of what the system looks like, sorry for the confusion
the only difference from this picture is that the mass moves up and down, and the vertical direction is the x axis 


#4
Mar711, 08:43 PM

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A mass attached to two horizontal springs, vertical motion
Do a freebody diagram for m in the supposed equilibrium position. What are the downward forces on m? What is the upward force? How can this be equilibrium? If m is on a horizontal surface, you can stretch the springs by an equal amount and achieve equilibrium. But when you turn it vertically, the springs will not be stretched equally. AM 


#5
Mar711, 09:51 PM

P: 70

i dont know, but were asked to just ignore gravity so the only forces acting on it when you displace the mass are the two forces of the springs, and no there isnt a horizontal surface



#6
Mar811, 06:45 AM

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Your equation is a differential equation that has a general solution of: [tex]x = A\sin{(\omega t + \phi)}[/tex] where [tex]\omega = \sqrt{2k/m}[/tex] and [itex]\phi[/itex] is a phase factor that depends on your choice of when t = 0. What is the value of A? AM 


#7
Mar811, 10:20 AM

P: 70

wait, wouldnt w= squareroot(2ksin(theta)/m)?
because the sin(theta) i got for the acceleration just depends on the initial angle the mass starts from when displaced, then x = Asin(omegat + phi) and im not sure what A would be 


#8
Mar811, 11:38 AM

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[tex]\frac{d^2x}{dt^2} = A\frac{d}{dt}(\omega(\cos{\omega t + \phi})) = A\omega^2(\sin{(\omega t + \phi})) =  \omega^2 x = a[/tex] You have found that: [tex]a = 2kx/m[/tex] So that means that: [tex]\omega^2 = 2k/m[/tex] AM 


#9
Mar811, 06:43 PM

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If the motion is perpendicular to the springs, the solution is quite a bit different.
In your figure this would be the case if the mass moved in a direction into and then out of the page  away from the viewer then toward the viewer. 


#10
Mar811, 10:23 PM

P: 70

oh so the equation for the acceleration that i got is for if the spring moves forwards and backwards, not up and down?
what do i do to make it for the mass going up and down? 


#11
Mar811, 11:19 PM

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If the mass moves a distance x from it's equilibrium position, then each spring stretches from 2L to √(4L^{2}+x^{2}).
The component of force in the xdirection is just x/√(4L^{2}+x^{2}) times the magnitude of the force. 


#12
Mar911, 05:28 AM

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a(x)=2kx/m where x = [itex]\Delta x[/itex]  the displacement from the equilibrium position in the vertical direction. AM 


#13
Mar911, 02:42 PM

P: 70

hmm im going to ask my professor if we should assume the springs stay the same length when we displace the mass, because the question says were displacing it by small distances
and for the sin(theta), well if you displace the mass then the x component of the spring force would include sin(theta) 


#14
Mar911, 03:52 PM

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AM 


#15
Mar911, 05:26 PM

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no the question doesnt say that the springs are vertical, and from the diagram i gave shows that the springs are horizontal, but the x direction is vertical



#16
Mar911, 10:21 PM

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When the mass is displaced a distance x from equilibrium the magnitude of the force exerted by each spring is: [tex]F_s=k\left(\sqrt{4L^2+x^2}\,\,L\right)[/tex] Take the component of force in the xdirection & multiply by 2 to get the magnitude of the net Force. The net force, F, is in the negative xdirection. [tex]F=2k\left(\sqrt{4L^2+x^2}\,\,L\right)\frac{x}{\sqrt{4L^2+x^2}}=2kx\left(1\frac{L}{\sqrt{4L^2+x^2}}\right)=2kx\left(1\frac{1}{\sqrt{4+(x/L)^2}}\right)[/tex] The power series expansion for [tex]\frac{1}{\sqrt{4+u^2}}[/tex] is [tex]\frac{1}{\sqrt{4+u^2}\,}= {{1}\over{2}}{{x^2}\over{16}}+{{3 x^4}\over{256}}{{5 x^6}\over{2048}}+\dots[/tex] . 


#17
Mar911, 11:25 PM

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The restoring force on m is the vertical component of the spring force when the mass is displaced a distance x. The spring force is constant in magnitude (F = k2L) but changes in direction. It is only the vertical component that you are concerned about. Let the angle that the springs make with the horizontal be [itex]\theta[/itex]. Since the springs are identical and have the same stretch and length, they provide the same force and make the same angle for a given displacement.: [tex]F_x = k(2L)sin\theta + k(2L)sin\theta = 4kLsin\theta[/tex] Since [itex]sin\theta = x/2L[/itex] the equation becomes: [tex]F_x = 2kx[/tex] So: [tex]a = 2kx/m[/tex] and, therefore the solution is the one I gave you above. AM 


#18
Mar1011, 07:35 AM

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My derivation come out to F = 2kx(1/2+...) ≈ kx.
Check my derivation. I'll look later. 


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